Section 1.6. Quotient Groups 47 (iii) Prove that −I is the only element in Q of order 2, and that all other elements M = I satisfy M 2 = −I. Conclude that Q has a unique subgroup of order 2, namely, −I , and it is the center of Q. ∗ 1.68. Prove that the elements of Q can be relabeled as ±1, ±i, ±j, ±k, where i2 = j2 = k2 = −1, ij = k, jk = i, ki = j, ij = −ji, ik = −ki, jk = −kj. ∗ 1.69. Prove that the quaternions Q and the dihedral group D8 are nonisomorphic groups of order 8. ∗ 1.70. Prove that A4 is the only subgroup of S4 of order 12. ∗ 1.71. (i) For every group G, show that the function Γ: G → Aut(G), given by g → γg (where γx is conjugation by g), is a homomorphism. (ii) Prove that ker Γ = Z(G) and im Γ = Inn(G) conclude that Inn(G) is a subgroup of Aut(G). (iii) Prove that Inn(G) Aut(G). Section 1.6. Quotient Groups The construction of the additive group of integers modulo m is the prototype of a more general way of building new groups, called quotient groups, from given groups. The homomorphism π : Z → Im, defined by π : a → [a], is surjective, so that Im is equal to im π. Thus, every element of Im has the form π(a) for some a ∈ Z, and π(a) + π(b) = π(a + b). This description of the additive group Im in terms of the additive group Z can be generalized to arbitrary, not necessarily abelian, groups. Suppose that f : G → H is a surjective homomorphism between groups G and H. Since f is surjective, each element of H has the form f(a) for some a ∈ G, and the operation in H is given by f(a)f(b) = f(ab), where a, b ∈ G. Now ker f is a normal subgroup of G, and the First Isomorphism Theorem will reconstruct H = im f and the surjective homomorphism f from G and ker f alone. We begin by introducing a binary operation on the set S(G) of all nonempty subsets of a group G. If X, Y ∈ S(G), define XY = {xy : x ∈ X and y ∈ Y }. This multiplication is associative: X(Y Z) is the set of all x(yz), where x ∈ X, y ∈ Y , and z ∈ Z, (XY )Z is the set of all such (xy)z, and these are the same because (xy)z = x(yz) for all x, y, z ∈ G. Thus, S(G) is a semigroup in fact, S(G) is a monoid, for {1}Y = {1 · y : y ∈ Y } = Y = Y {1}. An instance of this multiplication is the product of a one-point subset {a} and a subgroup K ⊆ G, which is the coset aK. As a second example, we show that if H is any subgroup of G, then HH = H.

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